On the functional-completeness of the Sheffer stroke

Question

I have seen functional-completeness (in regards to boolean functions) defined as:

A set X of truth-functions (of 2-valued logic) is functionally complete if and only if for each of the five defined classes, there is a member of X which does not belong to that class

With those 5 classes being:

Truth-Preserving

False-Preserving

Linear

Monotone

Self-Dual

So since | is functionally-complete, this means it is not any one of the 5 type of functions listed above, right?

But I am not quite seeing how it is not self-dual, it seems like it IS self-dual. Could someone go over the self-dual functions and show the sheffer stroke is not self-dual?

Answer 1

A Boolean operator $f(x_1, \dotsc, x_n)$ is self-dual iff $ f(\neg x_1, \dotsc, \neg x_n) = \neg f(x_1, \dotsc, x_n)$.

In the case of the Sheffer stroke (NAND), $(p|q) \equiv \neg (p \land q) \equiv (\neg p \lor \neg q)$, so $(\neg p | \neg q) \equiv (p\lor q)$, whereas $\neg (p|q) \equiv (p \land q)$. As $\lor$ and $\land$ are distinct Boolean functions, $|$ isn't self-dual.